Question

The position of a particle along x-axis at time t is given by $x=2+t-3t^2$. The displacement and the distance travelled in the interval $t=0$ to $t=1$ are respectively:

• A. $-2,2.16$
• B. $0,2$
• C. $2,2$
• D. None of these

Answer 1

The correct option is A $-2,2.16$

$x=2+t-3t^2⟶1$

diff. w.r.t. time

$\frac{dx}{dt}=1-6t⟶2$

again differentiate w.r.t. time

$\frac{d{x}^2}{d{t}^2}=-6$

Hence motion is retarding because

$\frac{d^{2}x}{d{t}^2}<0$

Let at t time velocity of particle is zero

$\frac{dx}{dt}=0=1-6t$

$\therefore t=\frac{1}{6}$

at $\therefore t=\frac{1}{6}$ velocity of particle is zero, then particle moves in back direction at $t=1sec$

Now, displacement$=$ final position $-$ initial position

$=x\left(1\right)-x\left(0\right)$

$=2+1-3-2$

$=-2m$

Distance$=\left|x\right(1)-x\left(\frac{1}{6}\right)|+|x\left(\frac{1}{6}\right)-x\left(0\right)|$

$=|2+1-3-(2+\frac{1}{6}-\frac{3}{36})|+|2+\frac{1}{6}-\frac{3}{36}-2|$

$=|-2-\frac{1}{6}+\frac{1}{12}|+|2+\frac{1}{6}-\frac{1}{12}-2|$

$=|-2-\frac{1}{12}|+\left|\frac{1}{12}\right|$$=2+\frac{1}{12}+\frac{1}{12}$

$=2+\frac{1}{6}=\frac{13}{6}=2.16m$